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1. In a particualr expt 200cm of aq KMnO4 of conc 0.05mol dm-3 were mixed with 50cm of ethandioc acid of conc 0.5 mol dm-3 and 80 cm3 of sulphuric acid.

Show by cac thast the starting conc of MnO4 ions was 3.03 X 10-2 mol dm-3??

its only 1 mark as well? how do u do it?
2. (Original post by Dust)
In a particualr expt 200cm of aq KMnO4 of conc 0.05mol dm-3 were mixed with 50cm of ethandioc acid of conc 0.5 mol dm-3 and 80 cm3 of sulphuric acid.

Show by cac thast the starting conc of MnO4 ions was 3.03 X 10-2 mol dm-3??

its only 1 mark as well? how do u do it?
Are you sure???
3. (Original post by Dust)
In a particualr expt 200cm of aq KMnO4 of conc 0.05mol dm-3 were mixed with 50cm of ethandioc acid of conc 0.5 mol dm-3 and 80 cm3 of sulphuric acid.

Show by cac thast the starting conc of MnO4 ions was 3.03 X 10-2 mol dm-3??

its only 1 mark as well? how do u do it?
OK, you have 0.05M MnO4- ions to start with in 200cm^3 volume so no. of mol = 0.05/1000 x 200 = 0.01 mol MnO4- ions.

You're then adding a total of 130cm^3 of other liquids giving a total volume of 330cm^3; so [MnO4-] = 0.01/330 x 1000 = 3.03 x 10^2 mol dm^-3.
4. so yer suggesting the 0.5M is useless?
5. (Original post by ResidentEvil)
so yer suggesting the 0.5M is useless?
Think so. It works without taking it into account. At a guess I'd reckon this is the first part of a long question and the 0.5M will become important in later parts.

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